Two particles P and Q are executing SHM across same straight line whose equations are given as `y_(P)=Asin (omegat+phi_(1)) and y_(Q)=Acos (omegat+phi_(2))`
An observer, at `t=0` observer the particle P at a distance `A//sqrt(2)` moving to the right from mean position O while particle Q at `(sqrt(3))/(2)` A moving to the left from mean position O as shown in figure. Then `phi_(2)-phi_(1)(i n rad)` ie equal to

Options

A. `5pi//6`
B. `3pi//4`
C. `5pi//2`
D. `7pi//12`

Answer

D

Explanation

`y_(p)=a sin (omegat+phi_(1)),`
For particle P, `t=0, y_(p)=(A)/(sqrt(2))`
`:. (A)/(sqrt(2))=A sin (0+phi_(1))=Asin phi_(1)`
or `sin phi_(1)=(1)/(sqrt(2)) or phi_(1)=(pi)/(4)rad`
`y_(Q)=A cos (omegat +phi_(2)),`
For particle `Q,t=0, y_(Q)=(-sqrt(3))/(2)A`
`:. -(sqrt(3))/(2)A=A cos (0 +phi_(2))=A cosphi_(2)`
or `cosphi_(2)=-(sqrt(3))/(2)=cos((5)/(6))pi or phi_(2)=(5pi)/(6)`
`:. phi_(2)-phi_(1)=(5pi)/(6)-(pi)/(4)=(7pi)/(12)rad`